Question

How do you find the asymptotes of $y=\dfrac{\left| x-1 \right|}{\left| x-2 \right|}$ ?

Answer 1

Now consider the given function. First we will check where the function is not defined. We know that function will not be defined when the denominator is 0, hence we get an asymptote with this condition. Now we will check the endpoints behavior of the graph. Hence we get the second asymptote.

Complete step by step solution:
Now we are given with a function $y=\dfrac{\left| x-1 \right|}{\left| x-2 \right|}$?
Asymptotes are values on the function which the function seems to approach but never reaches.
Hence asymptotes are the lines which represent the a value towards which the function seems to be approaching but never reaches.
Let us consider the function $\dfrac{1}{n}$ . Now we know that as $n\to \infty$ the function $\dfrac{1}{n}\to 0$ . Hence we have the function approaches 0 but never reaches 0 as $\dfrac{1}{n}=0$ is not possible for any natural number n.
Now consider the given example $y=\dfrac{\left| x-1 \right|}{\left| x-2 \right|}$
Now we know that the function is not defined on x = 2.
Hence we have asymptote at x = 2.
Now let us check the endpoints behavior of thee function
Now as $x\to \infty$ we have
$\begin{aligned} & \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left| x-1 \right|}{\left| x-2 \right|}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x-1}{x-2} \\\ & \Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{\left| x-1 \right|}{\left| x-2 \right|}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1-\dfrac{1}{x}}{1-\dfrac{2}{x}} \\\ \end{aligned}$
Hence as $x\to \infty$ we have $\dfrac{\left| x-1 \right|}{\left| x-2 \right|}\to 1$
Now similarly we have as $x\to -\infty$
$\begin{aligned} & \Rightarrow \underset{x\to -\infty }{\mathop{\lim }}\,\dfrac{\left| x-1 \right|}{\left| x-2 \right|}=\underset{x\to -\infty }{\mathop{\lim }}\,\dfrac{-x+1}{-x+2} \\\ & \Rightarrow \underset{x\to -\infty }{\mathop{\lim }}\,\dfrac{\left| x-1 \right|}{\left| x-2 \right|}=\underset{x\to -\infty }{\mathop{\lim }}\,\dfrac{-1+\dfrac{1}{x}}{-1+\dfrac{2}{x}} \\\ \end{aligned}$
Hence as $x\to -\infty$ we have $\dfrac{\left| x-1 \right|}{\left| x-2 \right|}\to 1$
Hence we can say y = 1 in asymptote.
Hence for the given function we have two asymptotes x = 2 and y = 1.

Note: Now note that we can have two times of asymptotes. If the line is parallel to x axis that means we have a horizontal asymptote and if the line is parallel to y axis then we have a vertical asymptote. Hence x = 2 is vertical asymptote and y = 1 is horizontal asymptote.